Generate random numbers following a normal distribution in C/C++

C++CRandomDistributionNormal Distribution

C++ Problem Overview


How can I easily generate random numbers following a normal distribution in C or C++?

I don't want any use of Boost.

I know that Knuth talks about this at length but I don't have his books at hand right now.

C++ Solutions


Solution 1 - C++

There are many methods to generate Gaussian-distributed numbers from a regular RNG.

The Box-Muller transform is commonly used. It correctly produces values with a normal distribution. The math is easy. You generate two (uniform) random numbers, and by applying an formula to them, you get two normally distributed random numbers. Return one, and save the other for the next request for a random number.

Solution 2 - C++

C++11

C++11 offers std::normal_distribution, which is the way I would go today.

C or older C++

Here are some solutions in order of ascending complexity:

  1. Add 12 uniform random numbers from 0 to 1 and subtract 6. This will match mean and standard deviation of a normal variable. An obvious drawback is that the range is limited to ±6 – unlike a true normal distribution.

  2. The Box-Muller transform. This is listed above, and is relatively simple to implement. If you need very precise samples, however, be aware that the Box-Muller transform combined with some uniform generators suffers from an anomaly called Neave Effect1.

  3. For best precision, I suggest drawing uniforms and applying the inverse cumulative normal distribution to arrive at normally distributed variates. Here is a very good algorithm for inverse cumulative normal distributions.

1. H. R. Neave, “On using the Box-Muller transformation with multiplicative congruential pseudorandom number generators,” Applied Statistics, 22, 92-97, 1973

Solution 3 - C++

A quick and easy method is just to sum a number of evenly distributed random numbers and take their average. See the Central Limit Theorem for a full explanation of why this works.

Solution 4 - C++

I created a C++ open source project for normally distributed random number generation benchmark.

It compares several algorithms, including

  • Central limit theorem method
  • Box-Muller transform
  • Marsaglia polar method
  • Ziggurat algorithm
  • Inverse transform sampling method.
  • cpp11random uses C++11 std::normal_distribution with std::minstd_rand (it is actually Box-Muller transform in clang).

The results of single-precision (float) version on iMac [email protected] , clang 6.1, 64-bit:

normaldistf

For correctness, the program verifies the mean, standard deviation, skewness and kurtosis of the samples. It was found that CLT method by summing 4, 8 or 16 uniform numbers do not have good kurtosis as the other methods.

Ziggurat algorithm has better performance than the others. However, it does not suitable for SIMD parallelism as it needs table lookup and branches. Box-Muller with SSE2/AVX instruction set is much faster (x1.79, x2.99) than non-SIMD version of ziggurat algorithm.

Therefore, I will suggest using Box-Muller for architecture with SIMD instruction sets, and may be ziggurat otherwise.


P.S. the benchmark uses a simplest LCG PRNG for generating uniform distributed random numbers. So it may not be sufficient for some applications. But the performance comparison should be fair because all implementations uses the same PRNG, so the benchmark mainly tests the performance of the transformation.

Solution 5 - C++

Here's a C++ example, based on some of the references. This is quick and dirty, you are better off not re-inventing and using the boost library.

#include "math.h" // for RAND, and rand
double sampleNormal() {
	double u = ((double) rand() / (RAND_MAX)) * 2 - 1;
	double v = ((double) rand() / (RAND_MAX)) * 2 - 1;
	double r = u * u + v * v;
	if (r == 0 || r > 1) return sampleNormal();
	double c = sqrt(-2 * log(r) / r);
	return u * c;
}

You can use a Q-Q plot to examine the results and see how well it approximates a real normal distribution (rank your samples 1..x, turn the ranks into proportions of total count of x ie. how many samples, get the z-values and plot them. An upwards straight line is the desired result).

Solution 6 - C++

Use std::tr1::normal_distribution.

The std::tr1 namespace is not a part of boost. It's the namespace that contains the library additions from the C++ Technical Report 1 and is available in up to date Microsoft compilers and gcc, independently of boost.

Solution 7 - C++

This is how you generate the samples on a modern C++ compiler.

#include <random>
...
std::mt19937 generator;
double mean = 0.0;
double stddev  = 1.0;
std::normal_distribution<double> normal(mean, stddev);
cerr << "Normal: " << normal(generator) << endl;

Solution 8 - C++

You can use the GSL. Some complete examples are given to demonstrate how to use it.

Solution 9 - C++

Have a look on: http://www.cplusplus.com/reference/random/normal_distribution/. It's the simplest way to produce normal distributions.

Solution 10 - C++

If you're using C++11, you can use std::normal_distribution:

#include <random>

std::default_random_engine generator;
std::normal_distribution<double> distribution(/*mean=*/0.0, /*stddev=*/1.0);

double randomNumber = distribution(generator);

There are many other distributions you can use to transform the output of the random number engine.

Solution 11 - C++

I've followed the definition of the PDF given in http://www.mathworks.com/help/stats/normal-distribution.html and came up with this:

const double DBL_EPS_COMP = 1 - DBL_EPSILON; // DBL_EPSILON is defined in <limits.h>.
inline double RandU() {
    return DBL_EPSILON + ((double) rand()/RAND_MAX);
}
inline double RandN2(double mu, double sigma) {
    return mu + (rand()%2 ? -1.0 : 1.0)*sigma*pow(-log(DBL_EPS_COMP*RandU()), 0.5);
}
inline double RandN() {
    return RandN2(0, 1.0);
}

It is maybe not the best approach, but it's quite simple.

Solution 12 - C++

There exists various algorithms for the inverse cumulative normal distribution. The most popular in quantitative finance are tested on http://chasethedevil.github.io/post/monte-carlo-inverse-cumulative-normal-distribution/

In my opinion, there is not much incentive to use something else than algorithm AS241 from Wichura: it is machine precision, reliable and fast. Bottlenecks are rarely in the Gaussian random number generation.

The top answer here advocates for Box-Müller, you should be aware that it has known deficiencies. I quote https://www.sciencedirect.com/science/article/pii/S0895717710005935: > in the literature, Box–Muller is sometimes regarded as slightly inferior, mainly for two reasons. First, if one applies the Box–Muller method to numbers from a bad linear congruential generator, the transformed numbers provide an extremely poor coverage of the space. Plots of transformed numbers with spiraling tails can be found in many books, most notably in the classic book of Ripley, who was probably the first to make this observation"

Solution 13 - C++

The comp.lang.c FAQ list shares three different ways to easily generate random numbers with a Gaussian distribution.

You may take a look of it: http://c-faq.com/lib/gaussian.html

Solution 14 - C++

Box-Muller implementation:

#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iostream>
using namespace std;
 // return a uniformly distributed random number
double RandomGenerator()
{
  return ( (double)(rand()) + 1. )/( (double)(RAND_MAX) + 1. );
}
 // return a normally distributed random number
double normalRandom()
{
  double y1=RandomGenerator();
  double y2=RandomGenerator();
  return cos(2*3.14*y2)*sqrt(-2.*log(y1));
}

int main(){
double sigma = 82.;
double Mi = 40.;
  for(int i=0;i<100;i++){
double x = normalRandom()*sigma+Mi;
    cout << " x = " << x << endl;
  }
  return 0;
}

Solution 15 - C++

  1. Graphically intuitive way you can generate Gaussian random numbers is by using something similar to the Monte Carlo method. You would generate a random point in a box around the Gaussian curve using your pseudo-random number generator in C. You can calculate if that point is inside or underneath the Gaussian distribution using the equation of the distribution. If that point is inside the Gaussian distribution, then you have got your Gaussian random number as the x value of the point.

This method isn't perfect because technically the Gaussian curve goes on towards infinity, and you couldn't create a box that approaches infinity in the x dimension. But the Guassian curve approaches 0 in the y dimension pretty fast so I wouldn't worry about that. The constraint of the size of your variables in C may be more of a limiting factor to your accuracy.

  1. Another way would be to use the Central Limit Theorem which states that when independent random variables are added, they form a normal distribution. Keeping this theorem in mind, you can approximate a Gaussian random number by adding a large amount of independent random variables.

These methods aren't the most practical, but that is to be expected when you don't want to use a preexisting library. Keep in mind this answer is coming from someone with little or no calculus or statistics experience.

Solution 16 - C++

Monte Carlo method The most intuitive way to do this would be to use a monte carlo method. Take a suitable range -X, +X. Larger values of X will result in a more accurate normal distribution, but takes longer to converge. a. Choose a random number z between -X to X. b. Keep with a probability of N(z, mean, variance) where N is the gaussian distribution. Drop otherwise and go back to step (a).

Solution 17 - C++

Take a look at what I found.

This library uses the Ziggurat algorithm.

Solution 18 - C++

Computer is deterministic device. There is no randomness in calculation. Moreover arithmetic device in CPU can evaluate summ over some finite set of integer numbers (performing evaluation in finite field) and finite set of real rational numbers. And also performed bitwise operations. Math take a deal with more great sets like [0.0, 1.0] with infinite number of points.

You can listen some wire inside of computer with some controller, but would it have uniform distributions? I don't know. But if assumed that it's signal is the the result of accumulate values huge amount of independent random variables then you will receive approximately normal distributed random variable (It was proved in Probability Theory)

There is exist algorithms called - pseudo random generator. As I feeled the purpose of pseudo random generator is to emulate randomness. And the criteria of goodnes is:

  • the empirical distribution is converged (in some sense - pointwise, uniform, L2) to theoretical
  • values that you receive from random generator are seemed to be idependent. Of course it's not true from 'real point of view', but we assume it's true.

One of the popular method - you can summ 12 i.r.v with uniform distributions....But to be honest during derivation Central Limit Theorem with helping of Fourier Transform, Taylor Series, it is neededed to have n->+inf assumptions couple times. So for example theoreticaly - Personally I don't undersand how people perform summ of 12 i.r.v. with uniform distribution.

I had probility theory in university. And particulary for me it is just a math question. In university I saw the following model:


double generateUniform(double a, double b)
{
  return uniformGen.generateReal(a, b);
}
        
double generateRelei(double sigma)
{
  return sigma * sqrt(-2 * log(1.0 - uniformGen.generateReal(0.0, 1.0 -kEps)));
}
double generateNorm(double m, double sigma)
{
  double y2 = generateUniform(0.0, 2 * kPi);
  double y1 = generateRelei(1.0);
  double x1 = y1 * cos(y2);
  return sigma*x1 + m;
}

Such way how todo it was just an example, I guess it exist another ways to implement it.

Provement that it is correct can be found in this book "Moscow, BMSTU, 2004: XVI Probability Theory, Example 6.12, p.246-247" of Krishchenko Alexander Petrovich ISBN 5-7038-2485-0

Unfortunately I don't know about existence of translation of this book into English.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionDamienView Question on Stackoverflow
Solution 1 - C++S.LottView Answer on Stackoverflow
Solution 2 - C++Peter G.View Answer on Stackoverflow
Solution 3 - C++Paul RView Answer on Stackoverflow
Solution 4 - C++Milo YipView Answer on Stackoverflow
Solution 5 - C++Pete855217View Answer on Stackoverflow
Solution 6 - C++JoeGView Answer on Stackoverflow
Solution 7 - C++PetterView Answer on Stackoverflow
Solution 8 - C++Denis ArnaudView Answer on Stackoverflow
Solution 9 - C++telcomView Answer on Stackoverflow
Solution 10 - C++Drew NoakesView Answer on Stackoverflow
Solution 11 - C++MJVCView Answer on Stackoverflow
Solution 12 - C++jherekView Answer on Stackoverflow
Solution 13 - C++DelganView Answer on Stackoverflow
Solution 14 - C++SysadminView Answer on Stackoverflow
Solution 15 - C++dan danView Answer on Stackoverflow
Solution 16 - C++JagatView Answer on Stackoverflow
Solution 17 - C++dwbritoView Answer on Stackoverflow
Solution 18 - C++Konstantin BurlachenkoView Answer on Stackoverflow